I want to prove that $$<a,b \ | \ aba^{-1}b^{-1},ab^{-1}ab>\cong \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$$ I already show $a^2=1$, but I don't make sure that $a\neq1$. How can I prove this?
Any help would appreciated. Thanks.
I want to prove that $$<a,b \ | \ aba^{-1}b^{-1},ab^{-1}ab>\cong \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$$ I already show $a^2=1$, but I don't make sure that $a\neq1$. How can I prove this?
Any help would appreciated. Thanks.
If $a=1$, then your presentation collapses to the simple presentation $\langle b\ |\ \ \rangle\cong\mathbb Z$.